Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $q = \dfrac{-2n - 10}{-n + 2} \div \dfrac{n^2 + 9n + 20}{-2n^2 + 4n} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-2n - 10}{-n + 2} \times \dfrac{-2n^2 + 4n}{n^2 + 9n + 20} $ First factor the quadratic. $q = \dfrac{-2n - 10}{-n + 2} \times \dfrac{-2n^2 + 4n}{(n + 5)(n + 4)} $ Then factor out any other terms. $q = \dfrac{-2(n + 5)}{-(n - 2)} \times \dfrac{-2n(n - 2)}{(n + 5)(n + 4)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -2(n + 5) \times -2n(n - 2) } { -(n - 2) \times (n + 5)(n + 4) } $ $q = \dfrac{ 4n(n + 5)(n - 2)}{ -(n - 2)(n + 5)(n + 4)} $ Notice that $(n - 2)$ and $(n + 5)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 4n\cancel{(n + 5)}(n - 2)}{ -(n - 2)\cancel{(n + 5)}(n + 4)} $ We are dividing by $n + 5$ , so $n + 5 \neq 0$ Therefore, $n \neq -5$ $q = \dfrac{ 4n\cancel{(n + 5)}\cancel{(n - 2)}}{ -\cancel{(n - 2)}\cancel{(n + 5)}(n + 4)} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $q = \dfrac{4n}{-(n + 4)} $ $q = \dfrac{-4n}{n + 4} ; \space n \neq -5 ; \space n \neq 2 $